BZOJ1128: [POI2008]Lam

因为互质有很多美妙方便的性质，所以推一下可以弄出答案的柿子
ansi=1ai∏nj=i+1aj−1aj
于是这个分子分母会非常大…要写高精
因为要最简，需要资瓷一下高精和单精取gcd，因为有(a,b)=(a,bModa)，写个高精模单精后直接gcd就行了
然后资瓷一下高精和单精的基本运算qwq

注意特判最后一个ai=1的位置前面直接输出0/1，处理的时候忽略掉否则会有/0的问题

code：

#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long longusing namespace std;int gcd(int a,int b){return !a?b:gcd(b%a,a);}const int maxn = 1100;const int maxl = 11000;int n;int a[maxn];struct A{    int a[maxl],len;}nowa,nowb;void output(A &x){for(int i=x.len-1;i>=0;i--) putchar('0'+x.a[i]);}void print() { output(nowa); putchar('/'); output(nowb); putchar('\n'); }int mod(A &x,int k){    ll re=0;    for(int i=x.len-1;i>=0;i--)        re=(re*10ll+x.a[i])%k;    return re;}int div(A &x,int k){    ll re=0ll;    for(int i=x.len-1;i>=0;i--)    {        re=re*10ll+x.a[i];        x.a[i]=re/k;        re=re%k;    }    while(!x.a[x.len-1]) x.len--;}int mul(A &x,int k){    for(int i=x.len-1;i>=0;i--)    {        ll tmp=(ll)x.a[i]*k;        if(tmp>9) x.a[i+1]+=tmp/10,tmp%=10;        x.a[i]=tmp;    }    for(int i=0;i<x.len;i++)        if(x.a[i]>9) x.a[i+1]+=x.a[i]/10,x.a[i]%=10;    while(x.a[x.len])    {        int &tmp=x.a[x.len];        x.a[x.len+1]+=tmp/10;        tmp%=10; x.len++;    }}int main(){    scanf("%d",&n);    int las=0;    for(int i=1;i<=n;i++)     {        scanf("%d",&a[i]);        if(a[i]==1) las=i-1;    }    for(int i=1;i<=las;i++) printf("0/1\n");    for(int i=1;i<=n-las;i++) a[i]=a[i+las]; n-=las;    nowa.len=nowb.len=1;    nowa.a[0]=nowb.a[0]=1;    for(int i=n;i>1;i--)    {        int tmp=mod(nowa,a[i]); tmp=gcd(tmp,a[i]);        int k=a[i]; if(tmp!=1) div(nowa,tmp),k/=tmp;        mul(nowb,k);        tmp=mod(nowb,a[i]-1); tmp=gcd(tmp,a[i]-1);        k=a[i]-1; if(tmp!=1) div(nowb,tmp),k/=tmp;        mul(nowa,k);    }    int tmp=mod(nowa,a[1]); tmp=gcd(tmp,a[1]);    int k=a[1]; if(tmp!=1) div(nowa,tmp),k/=tmp;    mul(nowb,k);    print();    for(int i=2;i<=n;i++)    {        k=a[i-1]; tmp=mod(nowb,k); tmp=gcd(tmp,k);        if(tmp!=1) k/=tmp,div(nowb,tmp);        mul(nowa,k);        k=a[i]-1; tmp=mod(nowa,k); tmp=gcd(tmp,k);        if(tmp!=1) k/=tmp,div(nowa,tmp);        mul(nowb,k);        print();    }    return 0;}